Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $p \neq 0$. $y = \dfrac{4p - 28}{p^3 - 12p^2 + 35p} \div \dfrac{p - 6}{4p^3 - 48p^2 + 144p} $
Answer: Dividing by an expression is the same as multiplying by its inverse. $y = \dfrac{4p - 28}{p^3 - 12p^2 + 35p} \times \dfrac{4p^3 - 48p^2 + 144p}{p - 6} $ First factor out any common factors. $y = \dfrac{4(p - 7)}{p(p^2 - 12p + 35)} \times \dfrac{4p(p^2 - 12p + 36)}{p - 6} $ Then factor the quadratic expressions. $y = \dfrac {4(p - 7)} {p(p - 7)(p - 5)} \times \dfrac {4p(p - 6)(p - 6)} {p - 6} $ Then multiply the two numerators and multiply the two denominators. $y = \dfrac {4(p - 7) \times 4p(p - 6)(p - 6) } { p(p - 7)(p - 5) \times (p - 6)} $ $y = \dfrac {16p(p - 6)(p - 6)(p - 7)} {p(p - 7)(p - 5)(p - 6)} $ Notice that $(p - 7)$ and $(p - 6)$ appear in both the numerator and denominator so we can cancel them. $y = \dfrac {16p(p - 6)(p - 6)\cancel{(p - 7)}} {p\cancel{(p - 7)}(p - 5)(p - 6)} $ We are dividing by $p - 7$ , so $p - 7 \neq 0$ Therefore, $p \neq 7$ $y = \dfrac {16p\cancel{(p - 6)}(p - 6)\cancel{(p - 7)}} {p\cancel{(p - 7)}(p - 5)\cancel{(p - 6)}} $ We are dividing by $p - 6$ , so $p - 6 \neq 0$ Therefore, $p \neq 6$ $y = \dfrac {16p(p - 6)} {p(p - 5)} $ $ y = \dfrac{16(p - 6)}{p - 5}; p \neq 7; p \neq 6 $